Integration by Observation

In integration by observation, we look at a product or a quotient to see a certain mathematical connection between its two parts.

Case 1



If the numerator of the fraction is the differential coefficient of the denominator, then the integral is ln|denominator|


Proof:

∫ f '(x) / f(x) dx
let u = f(x) => du/dx = f '(x) => du = f '(x)dx
∫ f '(x) / f(x) dx = ∫ 1/u du
= ln|u| + c
= ln|f(x)| +c

E.g.1

∫ cot x dx
= ∫ cos x/sin x
= ln|sin x| + c

E.g.2

∫ tan x dx
= ∫ sin x/cos x
= - ∫ -sin x/cos x
= - ln|cos x| + c
or 0 - ln|cos x| = ln 1 - ln|cos x| = ln|1/cos x| = ln|sec x| + c

E.g.3

∫ x+1/x2 + 2x + 3 dx
= 1/2 ∫ 2x+2/x2 + 2x + 3 dx
= 1/2 ln|x2 + 2x + 3| + c

E.g.4

∫ 1/tan x cos2x dx
= ∫ sec2 x/tan x dx
= ln|tan x| + c



Case 2



Please note that f '(x) is the differential coefficient of just the inner function, not the whole function.


Proof:

∫ f '(x) [f(x)]n dx
let u = f(x) => du/dx = f '(x) => du = f '(x) dx
∫ f '(x) / [f(x)]n dx = ∫ un du
= un+1/n+1 + c
= [f(x)]n+1/n+1 + c

E.g.1

∫ sec2 x tan6 x dx
f '(x) = sec2 x and f(x) = tan x
∫ sec2 x tan6 x dx
= tan7x / 7 + c

E.g.2

∫ (x + 1) (x2 + 2x -3) dx
= 1/2 ∫ (2x + 2) (x2 + 2x -3) dx
= (x2 + 2x -3)2/2 + c

E.g.3

∫ (x - 1) √ (x2 - 2x) dx
= 1/2 ∫ (2x - 2) √ (x2 - 2x) dx
= 1/2 (x2 - 2x)3/2/3/2 + c = (x2 - 2x)3/2/3 + c

E.g.3

∫ cos3x dx
= ∫ cos x cos2 x dx
= ∫ cos x (1 - sin2x) dx
= ∫ (cos x - cos x sin2x) dx
= sin x - sin3x / 3 + c



 

 

Recommended Reading

 

Maths is challenging; so is finding the right book. K A Stroud, in this book, cleverly managed to make all the major topics crystal clear with plenty of examples; popularity of the book speak for itself - 7th edition in print.

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